Class 10 Science · Chapter 11
Short answer:
Is chapter me NCERT ke saare 23 in-text questions (7 page-wise sets) aur saare 18 exercise questions step-by-step solve kiye gaye hain — NCERT ke exact order me. Coverage: electric current aur charge, potential difference, Ohm's law (V = IR), resistance ke factors aur resistivity (R = ρl/A), series aur parallel combinations, heating effect of current (H = I2Rt), aur electric power (P = VI = I2R = V2/R) with kWh commercial unit. Ye book ka sabse numerical-heavy chapter hai, isliye har numerical me given values, formula, substitution aur final answer with unit alag-alag dikhaya gaya hai. Saath me formula reference table, 6 common exam mistakes, board-style practice questions aur 6 FAQs bhi hain.
Electricity poore Class 10 Science ka sabse numerical-heavy chapter hai — aur iski achhi baat ye hai ki ismein sirf 5-6 formulas se 90% questions ban jaate hain. Baaki chapters me theory yaad karni padti hai, yahan aapko sirf ek cheez pakki karni hai: kaun sa formula kab lagta hai aur units kaise convert hote hain. Do line ka core idea ye hai — series me current har jagah same rehta hai, parallel me voltage har branch pe same rehta hai. Ye ek line yaad rahe to aadhe numericals bina soche ban jaate hain. Chapter ki shuruaat charge aur current se hoti hai, phir potential difference aata hai, uske baad Ohm's law (V = IR) — jo poore chapter ki reedh ki haddi hai. Fir resistance kis-kis cheez pe depend karta hai (length, area, material, temperature), phir series-parallel combinations, aur last me current ka heating effect aur power. Har section pichhle section pe khada hai, isliye beech me koi topic skip mat karna — warna aage ke numericals atak jaayenge. Aur ek warning pehle hi: is chapter me marks theory se nahi, calculation ki safai se aate hain. Unit likhna bhulna sabse aam aur sabse mehnga mistake hai.
Chapter 11 Summary — 5 Minute Revision
1. Electric Current aur Circuit
- Electric current = charge ke flow ka rate. Ek conductor ke kisi cross-section se per unit time jitna charge nikalta hai.
- Formula: I = Q/t. SI unit = ampere (A). 1 A = 1 C/s.
- 1 coulomb = 6.25 × 1018 electrons ka charge (kyunki 1 electron = 1.6 × 10-19 C).
- Conventional current ki direction electron flow ke ulti maani jaati hai — positive charge ka flow. Ye historical convention hai, isko bas maan lo.
- Electric circuit = current ka continuous aur closed path. Key band = circuit open = current zero.
- Current measure karne ka instrument = ammeter, hamesha circuit me series me jodte hain (kyunki uska resistance bahut kam hota hai).
2. Electric Potential aur Potential Difference
- Potential difference = do points ke beech unit charge le jaane me kiya gaya kaam. V = W/Q.
- SI unit = volt (V). 1 V = 1 J/C.
- "Do points ke beech 1 V potential difference hai" ka matlab: 1 C charge ko ek point se doosre point tak le jaane me 1 joule kaam karna padta hai.
- Potential difference banaye rakhne ka device = cell / battery.
- Potential difference measure karne ka instrument = voltmeter, hamesha parallel me jodte hain (uska resistance bahut zyada hota hai).
3. Ohm's Law
- Constant temperature par, ek metallic conductor me current uske across potential difference ke directly proportional hota hai.
- V ∝ I ⇒ V = IR, jahan R = resistance.
- V vs I ka graph ek straight line through origin hota hai; uska slope = R (agar V y-axis pe ho).
- SI unit of resistance = ohm (Ω). 1 Ω = 1 V/A.
- Rheostat = variable resistance ka device, circuit me current control karne ke liye.
4. Resistance kis pe depend karta hai
- Length (l) — R ∝ l. Lamba wire = zyada resistance.
- Area of cross-section (A) — R ∝ 1/A. Mota wire = kam resistance = current aasani se behta hai.
- Material — resistivity ρ se.
- Temperature — metals ka resistance temperature badhne pe badhta hai.
Sab milakar: R = ρl/A, jahan ρ (rho) = resistivity, unit Ω m.
- Resistivity material ka apna property hai — wire ki length ya motai badalne se ρ nahi badalta, sirf R badalta hai.
- Silver sabse achha conductor hai (ρ = 1.60 × 10-8 Ω m), phir copper, phir aluminium.
- Alloys (nichrome, manganin, constantan) ki resistivity apne constituent metals se kaafi zyada hoti hai aur ye high temperature pe jaldi oxidise nahi hote — isliye heating elements alloy ke bante hain.
| Material | Resistivity (Ω m) | Type |
|---|---|---|
| Silver | 1.60 × 10-8 | Best conductor |
| Copper | 1.62 × 10-8 | Conductor (wiring) |
| Aluminium | 2.63 × 10-8 | Conductor (transmission) |
| Iron | 10.0 × 10-8 | Conductor |
| Mercury | 94.0 × 10-8 | Weak conductor |
| Nichrome | ~100 × 10-8 | Alloy (heating element) |
| Rubber / Glass | 1013 – 1017 | Insulator |
↔ Table ko side me swipe karein
5. Series Combination
- Current har resistor me same rehta hai.
- Voltage baant jaata hai: V = V1 + V2 + V3.
- Rs = R1 + R2 + R3 — equivalent resistance sabse bade individual resistance se bhi zyada hoti hai.
- Nuksan: ek device fuse ho jaaye to poora circuit band; har device ko apna required current nahi milta.
6. Parallel Combination
- Voltage har branch pe same rehta hai.
- Current baant jaata hai: I = I1 + I2 + I3.
- 1/Rp = 1/R1 + 1/R2 + 1/R3 — equivalent resistance sabse chhote individual resistance se bhi kam hoti hai.
- Do resistors ke liye shortcut: Rp = R1R2/(R1 + R2) (product upon sum) — sirf DO ke liye, teen ke liye nahi.
- n barabar resistors R parallel me: Rp = R/n.
- Ghar ki wiring parallel me hoti hai — har appliance ko poora 220 V milta hai, aur ek band ho to baaki chalte rehte hain.
7. Heating Effect of Electric Current
- Current jab resistance se guzarta hai to electrical energy heat me convert hoti hai — Joule's law of heating.
- H = I2Rt (joule me), aur H = VIt = V2t/R bhi.
- Practical applications: electric iron, toaster, heater, geyser, electric bulb filament (tungsten, melting point ~3380°C), electric fuse.
- Heater ka cord glow nahi karta kyunki uska resistance heating element ke mukable bahut kam hai — series me current same hai, isliye H ∝ R.
8. Electric Power
- Power = rate of doing work / rate of energy consumption.
- P = VI = I2R = V2/R. SI unit = watt (W) = 1 J/s.
- 1 kW = 1000 W.
- Commercial unit of electrical energy = kilowatt-hour (kWh), jise "1 unit" bolte hain.
- 1 kWh = 1000 W × 3600 s = 3.6 × 106 J.
- Bulb pe "220 V, 100 W" likha hai ka matlab: 220 V pe chalane par wo 100 W consume karega. Uska resistance = V2/P = 484 Ω.
9. Ek line ka revision
| Yaad rakho | Kyun kaam aata hai |
|---|---|
| Series → current same | Har resistor pe alag-alag voltage nikalne ke liye V = IR |
| Parallel → voltage same | Har branch ka current alag nikalne ke liye I = V/R |
| Series → R badhta hai | Answer sabse bade R se bada aana chahiye |
| Parallel → R ghatta hai | Answer sabse chhote R se chhota aana chahiye — verification trick |
| Rated appliance → R = V2/P | Bulb/heater ke numericals ka pehla step |
↔ Table ko side me swipe karein
In-Text Questions — Solutions
Q1. What does an electric circuit mean?
Electric circuit ka matlab hai electric current ka continuous aur closed path. Ismein ek source (cell ya battery), connecting wires, ek key/switch aur koi device (bulb, resistor) hote hain.
Jab tak path closed hai, current behta rahega. Key kholte hi (circuit open) path toot jaata hai aur current turant zero ho jaata hai — beech me gap ho to electron flow nahi ho sakta.
Q2. Define the unit of current.
Current ka SI unit ampere (A) hai.
Agar kisi conductor ke cross-section se 1 coulomb charge 1 second me guzarta hai, to us conductor me behne wala current 1 ampere kehlata hai.
I = Q/t
1 A = 1 C / 1 s
Ampere ek badi unit hai — practical circuits me aksar milliampere (1 mA = 10-3 A) aur microampere (1 µA = 10-6 A) use hote hain.
Q3. Calculate the number of electrons constituting one coulomb of charge.
Given: total charge Q = 1 C, ek electron ka charge e = 1.6 × 10-19 C
Q = n × e ⇒ n = Q / e
n = 1 / (1.6 × 10-19)
n = 6.25 × 1018
Answer: 1 coulomb charge me 6.25 × 1018 electrons hote hain. (Ye ek pure number hai, iska koi unit nahi hota.)
Q4. Name a device that helps to maintain a potential difference across a conductor.
Cell ya battery (battery = do ya zyada cells ka combination).
Cell ke andar ka chemical reaction uske terminals ke beech potential difference banaye rakhta hai, jiski wajah se conductor me current continuously behta rehta hai. DC generator bhi yahi kaam karta hai.
Q5. What is meant by saying that the potential difference between two points is 1 V?
Iska matlab hai ki 1 coulomb charge ko ek point se doosre point tak le jaane me 1 joule kaam karna padta hai.
V = W/Q
1 volt = 1 joule / 1 coulomb
Yaani volt = joule per coulomb. Potential difference dar-asal per unit charge energy hai — isliye zyada voltage matlab har coulomb charge zyada energy leke chal raha hai.
Q6. How much energy is given to each coulomb of charge passing through a 6 V battery?
Given: V = 6 V, Q = 1 C
V = W/Q ⇒ W = V × Q
W = 6 × 1 = 6 J
Answer: 6 joule energy har coulomb charge ko milti hai.
Q7. On what factors does the resistance of a conductor depend?
Conductor ka resistance chaar cheezon pe depend karta hai:
- Length (l) — R ∝ l. Wire jitna lamba, resistance utna zyada.
- Area of cross-section (A) — R ∝ 1/A. Wire jitna mota, resistance utna kam.
- Material (resistivity ρ) — silver/copper ki ρ kam, nichrome ki bahut zyada.
- Temperature — metallic conductors ka resistance temperature badhne pe badhta hai.
R = ρl/A
Q8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Current thick (mote) wire me zyada aasani se behta hai.
R = ρl/A ⇒ R ∝ 1/A
Mote wire ka area of cross-section (A) zyada hota hai, isliye uska resistance (R) kam hota hai. Same source matlab same potential difference V, aur Ohm's law se:
I = V/R
R kam ⇒ I zyada. Isliye mota wire zyada current pass karta hai.
Q9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Given: R constant, naya potential difference V′ = V/2
Pehle: I = V/R
Baad me: I′ = V′/R = (V/2)/R = V/2R
I′ = I/2
Answer: Current bhi aadha (half) ho jaayega. Kyunki R constant hai, I aur V directly proportional hain.
Q10. Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Do wajah se alloy (jaise nichrome) use hoti hai:
- Alloy ki resistivity apne constituent pure metals se kaafi zyada hoti hai — zyada R matlab H = I2Rt se zyada heat, jo heating device ke liye zaroori hai.
- Alloy high temperature par jaldi oxidise (burn) nahi hoti, jabki pure metal red-hot hone par hawa me jal jaata hai. Isliye alloy coil lambi chalti hai.
Alloy ka melting point bhi zyada hota hai, isliye coil pighalti nahi.
Q11. Use the data in Table 11.2 to answer the following: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
(a) Resistivity kam = better conductor.
ρiron = 10.0 × 10-8 Ω m
ρmercury = 94.0 × 10-8 Ω m
Iron ki resistivity mercury se kaafi kam hai, isliye iron better conductor hai.
(b) Table me sabse kam resistivity silver ki hai (1.60 × 10-8 Ω m), isliye silver best conductor hai.
Q12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Circuit ek single closed loop hoga jisme sab kuch series me hai:
- Teen cells (2 V each) ek ke baad ek — positive terminal agle cell ke negative se — kul 6 V ki battery.
- Uske baad plug key (switch).
- Phir 5 Ω, 8 Ω aur 12 Ω resistors ek line me.
- Aakhri resistor se wire wapas battery ke doosre terminal par.
Symbols yaad rakho: cell = ek lambi patli line (+) aur ek chhoti moti line (−); battery = aise kai cells ki series; resistor = rectangle (box); plug key = do circles with a gap.
Total emf = 2 + 2 + 2 = 6 V
Rs = 5 + 8 + 12 = 25 Ω
Q13. Redraw the circuit of Question 12, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Connections: Ammeter ko circuit ke saath series me lagao (kahin bhi, kyunki series me current har jagah same hai). Voltmeter ko sirf 12 Ω resistor ke across parallel me lagao.
Given: V = 6 V, R1 = 5 Ω, R2 = 8 Ω, R3 = 12 Ω (series)
Rs = R1 + R2 + R3 = 5 + 8 + 12 = 25 Ω
I = V/Rs = 6 / 25 = 0.24 A
V3 = I × R3 = 0.24 × 12 = 2.88 V
Answer: Ammeter reading = 0.24 A, Voltmeter reading = 2.88 V.
Q14. Judge the equivalent resistance when the following are connected in parallel: (a) 1 Ω and 106 Ω, (b) 1 Ω and 103 Ω and 106 Ω.
(a) 1 Ω aur 106 Ω parallel me:
1/Rp = 1/1 + 1/106 = 1 + 0.000001 = 1.000001
Rp = 1/1.000001 ≈ 0.999999 Ω ≈ 1 Ω
(b) 1 Ω, 103 Ω aur 106 Ω parallel me:
1/Rp = 1/1 + 1/103 + 1/106 = 1 + 0.001 + 0.000001 = 1.001001
Rp = 1/1.001001 ≈ 0.999 Ω ≈ 1 Ω
Judgement: Dono cases me equivalent resistance lagbhag 1 Ω hai, yaani sabse chhote resistance se thoda sa kam. Ye parallel ka rule hai — Rp hamesha sabse chhote individual resistance se chhota hota hai, aur agar ek resistance bahut chhota ho to wahi answer ko decide karta hai.
Q15. An electric lamp of resistance 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Given: R1 = 100 Ω, R2 = 50 Ω, R3 = 500 Ω, V = 220 V (parallel)
1/Rp = 1/100 + 1/50 + 1/500
1/Rp = 5/500 + 10/500 + 1/500 = 16/500
Rp = 500/16 = 31.25 Ω
Electric iron ko utna hi current lena hai jitna teeno milkar lete hain, aur voltage bhi same 220 V hai — iska matlab uska resistance bhi wahi hoga:
Riron = 31.25 Ω
I = V/R = 220 / 31.25 = 7.04 A
Answer: Electric iron ka resistance = 31.25 Ω aur usme behne wala current = 7.04 A.
Q16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Parallel connection ke chaar bade fayde:
- Har device ko poora voltage milta hai (jaise 220 V), isliye har appliance apni rated power pe theek se chalta hai. Series me voltage baant jaata hai.
- Ek device fail ho jaaye to baaki chalte rehte hain — har branch ka apna alag path hai. Series me ek bulb fuse = sab band.
- Har device ko apni zaroorat ke hisaab se alag current milta hai — bulb ko kam, heater ko zyada. Series me sab ko ek hi current milta hai.
- Har device ka apna switch lag sakta hai, alag-alag on/off kar sakte hain.
Isi wajah se ghar ki poori domestic wiring parallel me hoti hai.
Q17. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
(a) 4 Ω ke liye: 3 Ω aur 6 Ω ko parallel me jodo, aur us combination ko 2 Ω ke saath series me.
1/R′ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2 ⇒ R′ = 2 Ω
Rtotal = 2 + R′ = 2 + 2 = 4 Ω ✓
(b) 1 Ω ke liye: teeno resistors ko parallel me jodo.
1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
Rp = 1 Ω ✓
Q18. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
(a) Highest resistance — sab coils ko series me jodo:
Rs = 4 + 8 + 12 + 24 = 48 Ω
(b) Lowest resistance — sab coils ko parallel me jodo:
1/Rp = 1/4 + 1/8 + 1/12 + 1/24
LCM = 24 ⇒ 1/Rp = 6/24 + 3/24 + 2/24 + 1/24 = 12/24 = 1/2
Rp = 2 Ω
Answer: Highest = 48 Ω (series), Lowest = 2 Ω (parallel).
Check: 2 Ω sabse chhote coil (4 Ω) se bhi kam hai — parallel ka rule sahi baith raha hai.
Q19. Why does the cord of an electric heater not glow while the heating element does?
Heater ka cord aur heating element series me hain, isliye dono me current same (I) behta hai.
H = I2Rt ⇒ I aur t same hone par H ∝ R
Heating element nichrome jaise alloy ka bana hota hai jiska resistance bahut zyada hai, jabki cord copper ka hota hai jiska resistance bahut kam hai.
Isliye element me itni heat paida hoti hai ki wo red-hot hokar glow karta hai, jabki cord me na ke barabar heat banti hai aur wo thanda rehta hai.
Q20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Given: Q = 96000 C, V = 50 V, t = 1 h = 3600 s
H = V × Q
H = 50 × 96000
H = 4,800,000 J = 4.8 × 106 J
Answer: 4.8 × 106 J
Note: Yahan time ki zaroorat calculation me nahi padi kyunki H = VQ direct de deta hai. Agar current chahiye ho to I = Q/t = 96000/3600 = 26.67 A.
Q21. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Given: R = 20 Ω, I = 5 A, t = 30 s
H = I2Rt
H = (5)2 × 20 × 30
H = 25 × 20 × 30 = 15,000 J
Answer: 15,000 J = 1.5 × 104 J
Q22. What determines the rate at which energy is delivered by a current?
Energy deliver hone ka rate hi electric power kehlata hai.
P = VI = I2R = V2/R
Yaani rate ko decide karte hain — circuit ka potential difference (V) aur usme behne wala current (I) (ya equivalently resistance R). SI unit = watt (W) = 1 J/s.
Q23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Given: I = 5 A, V = 220 V, t = 2 h
P = V × I = 220 × 5 = 1100 W = 1.1 kW
Energy E = P × t = 1.1 kW × 2 h = 2.2 kWh
Joule me: E = 1100 W × (2 × 3600 s) = 1100 × 7200 = 7.92 × 106 J
Answer: Power = 1100 W, Energy consumed = 2.2 kWh (= 7.92 × 106 J).
Poore Class 10 Science ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is — (a) 1/25 (b) 1/5 (c) 5 (d) 25
Answer: (d) 25
Wire ko 5 barabar tukdon me kaatne pe har tukde ki length 1/5 reh jaati hai, aur R ∝ l, isliye har tukde ka resistance:
Reach = R/5
Ab 5 aise barabar resistors parallel me:
R′ = (R/5) / 5 = R/25
R/R′ = R / (R/25) = 25
Q2. Which of the following terms does not represent electrical power in a circuit? — (a) I2R (b) IR2 (c) VI (d) V2/R
Answer: (b) IR2
Power ke valid roop hain:
P = VI P = I2R P = V2/R
IR2 in me se koi nahi hai — ye galat expression hai. (Yaad rakhne ka tareeka: I ke saath R ka square nahi, I ka square hota hai.)
Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be — (a) 100 W (b) 75 W (c) 50 W (d) 25 W
Answer: (d) 25 W
Step 1 — bulb ka resistance (ye constant rehta hai):
R = V2/P = (220)2 / 100 = 48400 / 100 = 484 Ω
Step 2 — 110 V par power:
P′ = V′2/R = (110)2 / 484 = 12100 / 484 = 25 W
Shortcut: voltage aadha hone pe power P ∝ V2 se one-fourth ho jaati hai → 100/4 = 25 W.
Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be — (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Answer: (c) 1:4
Maan lo har wire ka resistance R hai, aur dono cases me same voltage V aur same time t.
Series:
Rs = R + R = 2R
Hs = V2t / Rs = V2t / 2R
Parallel:
Rp = R/2
Hp = V2t / Rp = 2V2t / R
Ratio:
Hs : Hp = (V2t/2R) : (2V2t/R) = (1/2) : 2 = 1 : 4
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Voltmeter ko un dono points ke across parallel me joda jaata hai jinke beech ka potential difference maapna hai.
Kyun parallel? Voltmeter ka resistance bahut zyada (ideally infinite) hota hai. Parallel me lagane se wo circuit ka current apni taraf nahi kheenchta aur original current lagbhag nahi badalta — isliye reading sahi aati hai.
Compare: Ammeter hamesha series me lagta hai kyunki uska resistance bahut kam (ideally zero) hota hai.
Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Given: d = 0.5 mm ⇒ r = 0.25 mm = 2.5 × 10-4 m, ρ = 1.6 × 10-8 Ω m, R = 10 Ω
Step 1 — area of cross-section:
A = πr2 = 3.14 × (2.5 × 10-4)2
A = 3.14 × 6.25 × 10-8 = 1.9625 × 10-7 m2
Step 2 — length nikalo:
R = ρl/A ⇒ l = RA/ρ
l = (10 × 1.9625 × 10-7) / (1.6 × 10-8)
l = (1.9625 × 10-6) / (1.6 × 10-8) = 122.7 m (approx)
Step 3 — diameter double karne par:
A ∝ d2 ⇒ d double ⇒ A 4 guna
R ∝ 1/A ⇒ R′ = R/4 = 10/4 = 2.5 Ω
Answer: Length ≈ 122.7 m. Diameter double karne par resistance one-fourth yaani 2.5 Ω ho jaata hai.
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below. Plot a graph between V and I and calculate the resistance of that resistor. I (A): 0.5, 1.0, 2.0, 3.0, 4.0 V (V): 1.6, 3.4, 6.7, 10.2, 13.2
V ko y-axis pe aur I ko x-axis pe plot karne par ek straight line through the origin milti hai — yahi Ohm's law ko verify karta hai. Us line ka slope hi resistance hai.
Har point se V/I nikalte hain:
1.6/0.5 = 3.2 Ω
3.4/1.0 = 3.4 Ω
6.7/2.0 = 3.35 Ω
10.2/3.0 = 3.4 Ω
13.2/4.0 = 3.3 Ω
Average:
R = (3.2 + 3.4 + 3.35 + 3.4 + 3.3) / 5 = 16.65 / 5 = 3.33 Ω
Answer: Resistance ≈ 3.4 Ω (chhota-mota farak experimental error ki wajah se hai).
Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Given: V = 12 V, I = 2.5 mA = 2.5 × 10-3 A
R = V/I
R = 12 / (2.5 × 10-3)
R = 4800 Ω = 4.8 × 103 Ω = 4.8 kΩ
Answer: 4800 Ω (4.8 kΩ)
Warning: mA ko A me convert karna mat bhulna — warna answer 4.8 Ω aa jaayega jo 1000 guna galat hai.
Q9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Given: V = 9 V; series resistors 0.2, 0.3, 0.4, 0.5, 12 Ω
Rs = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
I = V/Rs = 9 / 13.4 = 0.67 A (approx)
Answer: 0.67 A
Series combination me current har resistor me same hota hai, isliye 12 Ω resistor me bhi 0.67 A hi behega.
Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Given: V = 220 V, I = 5 A, har resistor R = 176 Ω
Step 1 — required equivalent resistance:
Rp = V/I = 220 / 5 = 44 Ω
Step 2 — n barabar resistors parallel me:
Rp = R/n ⇒ n = R/Rp
n = 176 / 44 = 4
Answer: 4 resistors parallel me chahiye.
Q11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
(i) 9 Ω ke liye: Do resistors ko parallel me jodo, aur teesre ko unke saath series me.
Rparallel = (6 × 6)/(6 + 6) = 36/12 = 3 Ω
Rtotal = 3 + 6 = 9 Ω ✓
(ii) 4 Ω ke liye: Do resistors ko series me jodo, aur us combination ko teesre ke saath parallel me.
Rseries = 6 + 6 = 12 Ω
Rtotal = (12 × 6)/(12 + 6) = 72/18 = 4 Ω ✓
Q12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Given: V = 220 V, har bulb ki power = 10 W, max current = 5 A
Method 1 — current se:
Ek bulb ka current: I1 = P/V = 10/220 = 1/22 A
n = Imax / I1 = 5 / (1/22) = 5 × 22 = 110
Method 2 — resistance se (verification):
Ek bulb ka R = V2/P = (220)2/10 = 48400/10 = 4840 Ω
Required Rp = V/I = 220/5 = 44 Ω
n = 4840 / 44 = 110 ✓
Answer: 110 lamps
Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Given: V = 220 V, RA = RB = 24 Ω
Case 1 — ek coil akela:
I = V/R = 220 / 24 = 9.17 A (approx)
Case 2 — dono series me:
Rs = 24 + 24 = 48 Ω
I = 220 / 48 = 4.58 A (approx)
Case 3 — dono parallel me:
Rp = (24 × 24)/(24 + 24) = 576/48 = 12 Ω
I = 220 / 12 = 18.33 A (approx)
Answer: Separately 9.17 A, series me 4.58 A, parallel me 18.33 A. Yaani parallel setting pe oven sabse tez garam hoga.
Q14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
(i) Series circuit — 6 V, 1 Ω + 2 Ω:
Rs = 1 + 2 = 3 Ω
I = V/Rs = 6/3 = 2 A
P2Ω = I2R = (2)2 × 2 = 4 × 2 = 8 W
(ii) Parallel circuit — 4 V across 12 Ω and 2 Ω:
Parallel me har branch pe poora 4 V lagta hai, isliye 2 Ω ke across V = 4 V.
P2Ω = V2/R = (4)2 / 2 = 16/2 = 8 W
Answer: Dono cases me 2 Ω resistor 8 W power use karta hai — yaani power barabar hai (ratio 1:1).
Note: Part (ii) me 12 Ω wale branch ka koi role nahi hai kyunki parallel branches ek doosre ka voltage affect nahi karte.
Q15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Given: P1 = 100 W, P2 = 60 W, V = 220 V (parallel)
Parallel me total power add hoti hai:
Ptotal = 100 + 60 = 160 W
I = P/V = 160 / 220 = 0.727 A
Answer: ≈ 0.73 A
Alternative check: I1 = 100/220 = 0.4545 A, I2 = 60/220 = 0.2727 A, total = 0.727 A ✓
Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
TV set: P = 250 W, t = 1 h = 3600 s
ETV = P × t = 250 × 3600 = 9,00,000 J = 9 × 105 J
Toaster: P = 1200 W, t = 10 min = 600 s
Etoaster = 1200 × 600 = 7,20,000 J = 7.2 × 105 J
Answer: TV set zyada energy use karta hai (9 × 105 J vs 7.2 × 105 J).
Lesson: Zyada power wala device zaroori nahi zyada energy khaaye — time bhi utna hi zaroori hai. Energy = power × time.
Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Given: R = 8 Ω, I = 15 A
"Rate at which heat is developed" ka matlab hai power (heat per second) — isme time ki zaroorat nahi padti.
Rate = P = I2R
P = (15)2 × 8 = 225 × 8 = 1800 W
Answer: 1800 W = 1800 J/s
Agar total heat poocha jaata (2 h = 7200 s me): H = 1800 × 7200 = 1.296 × 107 J.
Q18. Explain the following. (a) Why is tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
(a) Tungsten filament: Tungsten ki resistivity bahut zyada hai aur uska melting point bahut ooncha (~3380°C) hai. Isliye wo bina pighle high temperature (~2500°C) tak garam hokar tez roshni deta hai. Koi aur aam metal itni garmi bardaasht nahi kar paata.
(b) Alloy in heating devices: Alloy (jaise nichrome) ki resistivity apne pure constituent metals se kaafi zyada hoti hai — H = I2Rt se zyada heat milti hai. Aur alloy high temperature par jaldi oxidise/burn nahi hoti, isliye coil lambi chalti hai.
(c) Series not used at home: Teen wajah — (i) series me total voltage baant jaata hai, isliye har appliance ko poora 220 V nahi milta aur wo theek se kaam nahi karta; (ii) ek device fuse hote hi poora circuit toot jaata hai aur sab band ho jaate hain; (iii) sab appliances ko ek hi current milta hai, jabki bulb aur heater ki current requirement alag-alag hai. Isliye ghar me parallel wiring hoti hai.
(d) Resistance vs area: R = ρl/A se R ∝ 1/A — area of cross-section badhne par resistance ghatta hai (inversely proportional). Area double = resistance aadha; diameter double = area 4 guna = resistance one-fourth.
(e) Copper aur aluminium: In dono ki resistivity bahut kam hai, isliye transmission lines me energy loss (I2Rt) kam hota hai. Saath hi ye ductile hain (taar ban jaate hain), mazboot hain aur silver ke mukable bahut saste hain.
Important Equations — Ek Nazar Me
| Quantity | Formula | SI Unit | Note |
|---|---|---|---|
| Electric current | I = Q/t | ampere (A) | 1 A = 1 C/s. Ammeter series me. |
| Charge ↔ electrons | Q = n × e | coulomb (C) | e = 1.6 × 10-19 C; 1 C = 6.25 × 1018 electrons |
| Potential difference | V = W/Q | volt (V) | 1 V = 1 J/C. Voltmeter parallel me. |
| Ohm's law | V = IR | — | Constant temperature par hi valid |
| Resistance | R = V/I | ohm (Ω) | 1 Ω = 1 V/A; V-I graph ka slope |
| Resistivity relation | R = ρl/A | ρ me Ω m | R ∝ l, R ∝ 1/A; ρ material ka property |
| Area (round wire) | A = πr2 = πd2/4 | m2 | Diameter ko radius me convert karna mat bhulna |
| Series resistance | Rs = R1 + R2 + R3 | Ω | Current same; Rs sabse bade R se bada |
| Parallel resistance | 1/Rp = 1/R1 + 1/R2 + 1/R3 | Ω | Voltage same; Rp sabse chhote R se chhota |
| Parallel (2 resistors) | Rp = R1R2/(R1+R2) | Ω | Product upon sum — sirf DO resistors ke liye |
| n equal resistors parallel | Rp = R/n | Ω | Bahut fast shortcut |
| Heat produced (Joule) | H = I2Rt = VIt = V2t/R | joule (J) | t hamesha seconds me |
| Electric power | P = VI = I2R = V2/R | watt (W) | 1 W = 1 J/s; 1 kW = 1000 W |
| Resistance of rated appliance | R = V2/P | Ω | "220 V, 100 W" wale bulb ka pehla step |
| Electrical energy | E = P × t | J ya kWh | kWh ke liye P kilowatt me aur t hours me |
| Commercial unit | 1 kWh = 3.6 × 106 J | kWh | "1 unit" of electricity |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Parallel resistances ko seedha jod dena. Sabse aam aur sabse mehnga mistake. 6 Ω aur 3 Ω parallel me 9 Ω nahi, 2 Ω dete hain. Yaad rakho: parallel me equivalent resistance hamesha sabse chhote individual resistance se bhi chhota hota hai. Answer likhne se pehle ye check zaroor karo — agar answer sabse chhote R se bada aa raha hai to calculation galat hai.
- 1/Rp nikaal kar ulta karna bhool jaana. Formula 1/Rp = 1/R1 + 1/R2 deta hai 1/Rp, khud Rp nahi. Jaise 1/Rp = 16/500 nikla to answer 16/500 nahi, 500/16 = 31.25 Ω hai. Har baar last step me reciprocal lena — ye ek step bhoolne se poora numerical zero ho jaata hai.
- Unit conversion me lapraahi — mA, kW, minutes, kWh. 2.5 mA ko 2.5 A maan lena answer ko 1000 guna galat kar deta hai. Rules: current ko ampere me, time ko seconds me (jab joule chahiye), power ko watt me lao. Sirf kWh nikaalte waqt power kilowatt me aur time hours me rakho. Aur 1 kWh = 3.6 × 106 J — 3600 nahi.
- P = I2R aur P = V2/R me confuse ho jaana. Dhyan do: current ke saath R multiply hota hai (P = I2R), voltage ke saath R se divide hota hai (P = V2/R). Ulta lagane se answer bilkul badal jaata hai. Simple rule: jo quantity di gayi hai (I ya V), usi wala formula uthao — beech me R nikalne ki zaroorat hi nahi padegi.
- Rated bulb ka resistance badalna maan lena. "220 V, 100 W" wala bulb 110 V pe chalane par uski power 50 W nahi, 25 W hoti hai. Kyunki bulb ka resistance constant rehta hai (R = V2/P = 484 Ω), aur P ∝ V2. Voltage aadha ⇒ power one-fourth. Isi tarah har rated-appliance numerical me pehla step hamesha R = V2/P nikalna hai.
- Diameter ko radius samajh lena aur area me square bhoolna. R = ρl/A me A = πr2 hai, aur r = d/2. 0.5 mm diameter ka matlab r = 0.25 mm = 2.5 × 10-4 m — 0.5 × 10-3 m nahi. Iske saath ye bhi yaad rakho: diameter double karne par area 4 guna hota hai (2 guna nahi), isliye resistance one-fourth hota hai, aadha nahi.
Board-Style Important Questions
- 1 mark: Define resistivity of a material. Write its SI unit. State whether it depends on the length of the conductor.
- 2 marks: An electric bulb is rated 12 V and 24 W. Calculate (i) the resistance of its filament and (ii) the current drawn by it when operated at its rated voltage.
- 3 marks: Two resistors of 6 Ω and 12 Ω are connected in parallel across a 12 V battery. Calculate (i) the equivalent resistance, (ii) the current through each resistor, and (iii) the total current drawn from the battery.
- 3 marks: State Joule's law of heating and derive the expression H = I2Rt. Explain why the connecting wires of an electric heater do not become red-hot while its element does.
- 5 marks: (a) State Ohm's law and draw the V–I graph for a metallic conductor, explaining what its slope represents. (b) An electric iron of 1100 W and a bulb of 100 W are used for 4 hours a day on a 220 V supply. Calculate the total electrical energy consumed in 30 days in kWh, and the current drawn by the iron.
Quick Quiz — Score Check Karein
Q1. Electric current ki definition kya hai aur uska formula?
Q2. 1 coulomb charge me kitne electrons hote hain (e = 1.6 × 10⁻¹⁹ C)?
Q3. Ammeter aur voltmeter circuit me kaise jode jaate hain?
Q4. Ohm's law ke mutabik, ek metallic conductor me constant temperature par V aur I ka relation kya hota hai?
Q5. Ek conductor ki length double kar di jaaye aur area same rahe, to uska resistance kaise badlega (R = ρl/A)?
Q6. Heating elements (jaise toaster, iron) nichrome jaisi alloy se kyun bante hain?
Q7. Series combination me 5Ω, 10Ω aur 15Ω resistors jode gaye hain. Equivalent resistance kya hoga?
Q8. Do resistors 4Ω aur 4Ω parallel me jode gaye hain. Equivalent resistance kya hoga (Rp = R1R2/(R1+R2))?
Q9. Ghar ki wiring series me na hokar parallel me kyun ki jaati hai?
Q10. Ek bulb par '220 V, 100 W' likha hai. Iska resistance kitna hoga (R = V²/P)?
Aksar Poochhe Jaane Wale Sawaal
Series aur parallel me kaunsa yaad rakhna sabse zaroori hai?
Sirf do lines: series me current har jagah same rehta hai aur resistances jud jaate hain (R_s = R1 + R2). Parallel me voltage har branch pe same rehta hai aur reciprocals jud jaate hain (1/Rp = 1/R1 + 1/R2). Aadhe se zyada numericals in do lines se hi ban jaate hain.
Answer sahi hai ya nahi, ye bina dobara solve kiye kaise check karun?
Ek quick sanity check: series ka equivalent resistance hamesha sabse bade individual resistance se bada aana chahiye, aur parallel ka hamesha sabse chhote se chhota. Agar aisa nahi hai to kahin calculation galat hai. Power aur current ke answers me unit bhi zaroor likho — bina unit ke poore marks nahi milte.
kWh aur kW me kya farak hai?
kW power ki unit hai (rate — kitni tezi se energy use ho rahi hai), aur kWh energy ki unit hai (total kitni energy use hui). 1 kWh = 1 kW ka device 1 ghante chalane pe consume hui energy = 3.6 x 10^6 joule. Bijli ka bill kWh (units) me aata hai, kW me nahi.
Ammeter series me aur voltmeter parallel me hi kyun lagta hai?
Ammeter ka resistance bahut kam (ideally zero) hota hai, isliye series me lagne par wo circuit ka current nahi badalta. Voltmeter ka resistance bahut zyada (ideally infinite) hota hai, isliye parallel me lagne par wo apni taraf current nahi kheenchta. Agar dono ulte lagayen to reading galat aayegi aur ammeter parallel me lagane se to short circuit ho jaayega.
Heating elements alloy ke aur bulb filament tungsten ka — dono me kya logic hai?
Dono me common baat hai high resistance + high melting point. Alloy (nichrome) ki resistivity pure metal se zyada hai aur wo garam hone par oxidise nahi hoti, isliye toaster/iron me chalti hai. Tungsten ka melting point bahut ooncha (~3380 C) hai, isliye wo 2500 C tak garam hokar bina pighle roshni de sakta hai.
Is chapter ke numericals me sabse zyada marks kahan katte hain?
Teen jagah — unit likhna bhool jaana, unit conversion galat karna (mA ko A me na badalna, minutes ko seconds me na badalna), aur parallel me reciprocal ulta karna bhool jaana. Har numerical me given values alag likho, formula likho, phir substitute karo. Step-wise likhne se calculation galat hone pe bhi method ke marks mil jaate hain.
Class 10 Science — Saare Chapters
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Class 10 Science ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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